Thom Spaces Normal Data and Collapse Maps — Examples

The disk bundle is $B\times [-1,1]$ and the sphere bundle is $B\times \{-1,1\}$. Collapsing both boundary copies to the one Thom basepoint gives $B_{+}\wedge ([-1,1]\mathrm{/}\{-1,1\})$.

The interval quotient is the based circle, so the result is $B_{+}\wedge S^1=\mathrm{\Sigma }{B}_{+}$. For a point base this is $S^1$; for empty base it is a point. The trivialization records which fiber direction is positive, even though the underlying homeomorphism type does not remember its sign.

The disk bundle $D(L)=[0,1]\times [-1,1]\mathrm{/}((0,t)\sim (1,-t))$ is a Möbius band and its sphere bundle $S(L)$ is its single boundary circle. Realize ${\mathbb{R}\mathbb{P}}^2$ as the disk with antipodal boundary points identified. Removing from it a smaller concentric open disk leaves the closed annulus with its outer circle antipodally identified, which is again a Möbius band: the outer identification reverses the inward transverse direction on one traversal, which is the defining twist. Since the removed disk is complementary to that Möbius band, ${\mathbb{R}\mathbb{P}}^2$ is obtained from $D(L)$ by attaching a closed disk along $\mathrm{\partial }D(L)$, and collapsing that attached disk turns the pushout into $D(L)\mathrm{/}\mathrm{\partial }D(L)$, which is $\mathrm{Th}(L)$ by [F1].

On a disk pair $D_1\subset D_2$ of concentric Euclidean disks, the radial map that sends $D_1$ to the center of $D_2$ and rescales the annulus $D_2\setminus D_1$ onto $D_2$ minus its center, while fixing the complement of $D_2$, is continuous, injective off $D_1$, and onto; it therefore descends to a continuous bijection $D_2\mathrm{/}{D}_1\to D_2$ from the compact quotient to the Hausdorff disk, hence a homeomorphism. Applying this inside the projective plane to a disk containing the attached disk exhibits $\mathrm{Th}(L)\cong {\mathbb{R}\mathbb{P}}^2\mathrm{/}{D}^2\cong {\mathbb{R}\mathbb{P}}^2$ with the basepoint corresponding to the centre of the complementary disk. The trivial line over $S^1$ instead has Thom space $\mathrm{\Sigma }(S_{+}^1)$ by [F2]: its two boundary circles are collapsed to the same basepoint, whereas the unreduced suspension $\mathrm{\Sigma }{S}^1\cong S^2$ keeps its two suspension points distinct.

The chart $\mathrm{\Phi }(x,t)=(\sqrt{1-t^2}x,t)$ is a diffeomorphism onto the band $\mathrm{\mid }t\mathrm{\mid }<1$. Its derivative in the normal direction at $t=0$ is $(0,1)$, so it respects exactly the specified framing. The metric disk of radius $a$ is its closed band, and [F1] gives the displayed formula.

The two band boundaries go to the circle basepoint, so the outside constant formula pastes continuously. By [F2] its target is $S_{+}^n\wedge S^1$. The continuous based map $S_{+}^n\to S^0$ sends the whole equator to the nonbasepoint; smashing it with the identity produces the claimed sphere-valued framed collapse. For $n=0$ the same formula has two band components and remains valid.

Both [F1] and [F2] use the same disk/sphere pair and the same fiber generators, so Thom uniqueness identifies their classes: the interface class of [F1] is the normalized class fixed by [F2]. Applying the relative-to-absolute map $j^{\ast }$ and then the zero-section map $s^{\ast }$ gives exactly $e(E)$, rather than an ill-typed direct pullback of a relative class.

For rank zero with standard unit orientation $u_E=1$ and $j^{\ast },s^{\ast }$ are identities, so $e(0_B)=1$; a different supplied orientation $o$ gives $e(0_B,o)=o$, the componentwise unit. For a positive-rank trivial bundle, choose a continuous unit-length section $s_1$ of the disk bundle (normalize a nowhere-zero section given by the supplied trivialization). The sections $s_t=t{s}_1$, $0\le t\le 1$, are homotopic in the disk bundle, so they pull $j^{\ast }{u}_E$ back to the same class; at $t=1$ the section factors through $S(E)$, where $i^{\ast }{j}^{\ast }{u}_E=0$ by exactness of the pair sequence. Hence the Euler class of a positive-rank trivial bundle is zero. Mod two the same definition requires no chosen orientation, and AC is inherited from the general AT suppliers.

At the point its tangent space is zero, so [F1] gives the normal fibers $\mathbb{R}$ and ${\mathbb{R}}^2$. These bundles have different ranks and are not isomorphic; rank is preserved by a fiberwise linear isomorphism.

Their Thom spaces are $S^1$ and $S^2$ by [F2]. They are not homeomorphic: removing any point from $S^1$ gives an open interval, and removing any further point disconnects it; removing a point from $S^2$ gives ${\mathbb{R}}^2$, which remains path connected after removing any further point (polygonal paths may be detoured around that point). A putative sphere homeomorphism would preserve these deletion properties. Nevertheless adding one trivial line to the first normal fiber gives the second and suspends its Thom sphere, in accordance with [F3]. This explicitly exhibits why only the stable normal class is intrinsic.